A bar weighing m = 3 kg rests on an inclined plane located at an angle α = 30 ° to the horizon.

A bar weighing m = 3 kg rests on an inclined plane located at an angle α = 30 ° to the horizon. The coefficient of friction between the bar and the plane is μ = 0.1. The bar is imparted with a short impact with a velocity u = 9 m / s directed upward along the plane. To what height h relative to the initial position will the bar rise?

A bar weighing m = 3 kg rests on an inclined plane located at an angle α = 30 ° to the horizon. The coefficient of friction between the bar and the plane is μ = 0.1. The bar is imparted with a short blow with a velocity u = 9 m / s directed upward along the plane. To what height h relative to the initial position will the bar rise?
m = 3 k.
∠α = 30 °.
μ = 0.1.
V = 9 m / s.
g = 9.8 m / s ^ 2.
h -?
According to the law of conservation of energy: the kinetic energy of the Ek bar is converted into potential energy En and into the work of the friction force Atr.
Ek = En + Atr.
The kinetic energy Ek is determined by the formula: Ek = m * V ^ 2/2.
Potential energy En is determined by the formula: En = m * g * h.
The work of the friction force Atr is determined by the formula: Atr = Ftr * S.
Ftr = μ * N, where μ is the coefficient of friction between the bar and the plane, N is the support reaction.
N = m * g * cosα.
Ftr = μ * m * g * cosα.
Atr = μ * m * g * cosα * S.
cosα = h / S.
Atr = μ * m * g * h.
m * V ^ 2/2 = m * g * h + μ * m * g * h.
V ^ 2/2 = (g + μ * g) * h.
h = V ^ 2/2 * (g + μ * g).
h = (9 m / s) ^ 2/2 * (9.8 m / s ^ 2 + 0.1 * 9.8 m / s ^ 2) = 1.878 m.
Answer: the body will rise to a height of h = 1,878 m.